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Question

If y=x(logx)log(logx), then dydx is

A
yx((lnxx1)+2lnxln(lnx))
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B
yx(logx)log(logx)(2log(logx)+1)
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C
yxlnx[(lnx)2+2ln(lnx)]
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D
yxlogylogx[2log(logx)+1]
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Solution

The correct options are
B yx(logx)log(logx)(2log(logx)+1)
D yxlogylogx[2log(logx)+1]
y=x(logx)log(logx)

Taking log of both sides, we get

logy=(logx)(logx)log(logx) ... (1)

Taking log of both sides, we get

log(logy)=log(logx)+log(logx)log(logx)

Differentiating w.r.t. x, we get

1logy.1ydydx=1xlogx+2log(logx)logx1x

=2log(logx)+1xlogx

or dydx=yx.logylogx[2log(logx)+1]

Substituting the value of y from (1), we get

dydx=yx(logx)log(logx)(2log(logx)+1)

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